Problem: $f(x) = \begin{cases} -6 & \text{if } x = 0 \\ -x^{2}-3 & \text{otherwise} \end{cases}$ What is the range of $f(x)$ ?
Answer: First consider the behavior for $x \ne 0$ Consider the range of $-x^{2}$ The range of $x^2$ is $\{\, y \mid y \ge 0 \,\}$ Multiplying by $-1$ flips the range to $\{\, y \mid y \le 0 \,\}$ To get $-x^{2}-3$ , we subtract $3$ So the range becomes: $\{\, y \mid y ≤ -3 \,\}$ If $x = 0$ , then $f(x) = -6$ , which eliminates $f(x) = -3$ from the range. The new range is $\{\, y \mid y < -3 \,\}$.